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12:00 AM
Ah, this is one of the standard bars and stripes applications. But I'm not doing any more combinatorics ;P
 
@TedShifrin Actually I think I have a nice way to explain what a fibration is. I'm basically gonna give the definition of a fibration really quickly, and since there's a theorem that says for any continuous map $p : E \to B$ we can turn this in a fibration, I'll say "You don't need to worry too much about the definition, just remember fibrations exist in abundance since we can turn any continuous map into one"
 
@Ted when I give my bachelor's presentation I will have given 8 seminar talks and will have TAed 4 times, so I should be okay
 
I'm not worried about you, @Mathein, as usual.
@Perturb: You just wasted a minute of your 13.
Be as concrete as possible.
You should not do most general statements in a short talk.
But, lucky for you, I have some stuff to do in the kitchen, so I'm disappearing for a bit.
 
but do you have enough time to explain the Hopf fibration in 13 minutes? That seems like the easiest example except for covering maps
 
@MatheinBoulomenos I won't have time to cover that unfortunately :(
How many people usually present their bachelor's thesis at your uni @Mathein?
 
12:05 AM
@Perturbative everyone
it's required
we actually call it "defending" the bachelor's thesis
 
How many of you are in your final year of your bachelor's usually in that case?
@TedShifrin Thanks for the advice, I'll try to keep things as concrete as possible
 
no idea. I don't usually pick up on what the students are doing who don't specialize in similar ares as I do
 
Ah okay
 
12:39 AM
False theorem: Inverses of bijective smooth functions are smooth. Counterexample: $\Bbb{R} \to \Bbb{R}$ given by $x \mapsto x^3$. What is the smooth structure on $\Bbb{R}$? I.e., what are charts/maps?
 
@user193319 that's backwards
 
What do you mean?
 
you want to do calculus on subsets of $\Bbb R^n$ first
and then extend it to manifolds
 
$x \mapsto x^3$ is smooth but $x \mapsto x^{1/3}$ isn't smooth.
 
smooth functions are defined generally for manifolds but firstly for subsets of $\Bbb R^n$
smooth just means infinitely differentiable
what I mean by backwards is "pedagogically backwards"
 
12:42 AM
hmm...I see...
 
@Leaky: What are you talking about?
@user193319: Yes, of course, you're right.
 
I thought you'd agree with me @Ted
 
Standard differentiable structure on $\Bbb R$.
 
I reserve the right to smack anyone, @Leaky.
14
 
12:46 AM
Ted is an equal-opportunity smacker.
 
the trouble with a nice relaxing nap: it's hard to muster up much motivation after
 
it's hard to be motivated before or after nap
 
I am not a good napper.
 
I hate getting up after only 4 hours of sleep. (less that 4 hours and I don't really feel that I fell asleep at all)
 
Even in college I think I managed more than 4 .... probably 6 on average.
But that was centuries ago, so who knows.
 
12:52 AM
I need to strongly resist the impulse to miss sleep now. I don't think I bounce back from it as well as I used to.
 
Wait 'til you're actually old, @Semiclassic.
 
I speak from a position of authority :D
 
I'm actually almost 30. Five years older than graduate students going for their doctorate. Spent too much time on other vocations where I wasn't suited.
 
It's OK, @Rithaniel. You're more motivated, for sure.
I got a Facebook message from an undergraduate at UGA who's taking "my" course and thanked me for my videos. Turns out he's your age. I have no idea what he was up to before.
 
1:01 AM
Yeah, I can agree with that. People seem apathetic more than anything.
 
I've noticed that more and more over the years. More students in school following their parents' orders/dreams than their own.
 
I've always been good at math, though. Should have gone for it from the beginning, instead of that stint where I attempted to be a chef.
 
Oh wow. That would have been my alternative career.
But it's more fun having it as a hobby and making friends happy, rather than trying to be a line cook for the masses.
My body would not have survived trying to run a restaurant.
We should cook sometime :)
 
You would have probably done better than me as a chef. I'm all about the simple recipes. Like, I made a burrito mix this week for my lunches. It's delicious (perhaps a bit too salty) but has no elegance to it. It's all brute force.
I tried going to culinary arts classes, but they kind of expected a higher level of entry proficiency than I had.
 
Oh, OK, I'm more snobby. French, Italian, Asian :P
Anyhow, you can have a better living with math/stat/computer science.
 
1:08 AM
Snobby can be good. It's means you have some finesse with the topic.
I'd need to start out with some bridging recipes before I could even consider the snobby stuff, but I'd be open to trying.
 
Just made pesto for tonight — totally simple.
These days you can use all sorts of things — basil, cilantro, arugula ... pine nuts, walnuts, whatever ...
Did you ever settle that element of finite order thing? Was the hypothesis that the gcd of the $n_i$ was not $1$?
 
Oh, I actually asked the professor. He recommended looking at the kernel of the homomorphism the hypothetical group would be the free group image of.
The hypothesis was that $G$ was a finitely generated abelian group where every element has infinite order.
 
I can give a counterexample. Indeed, I already did.
With that hypothesis, what's the conclusion? That the only solution to $\sum n_ig_i = 0$ is the trivial solution? That's clearly garbage.
You can have an extraneous generator and just set it equal to $0$.
 
The conclusion is that $G$ is free abelian, which is equivalent to saying that it has a nonempty basis, but my approach was not going to work because we don't know that the hypothetical generating subset will necessarily be a basis.
 
Right. You need a minimal generating set.
OK, now I believe the statement :)
 
1:16 AM
That was my take away from working on it with you and others, yesterday.
Right now I'm grading Calc 3 papers, though.
 
Yum, my favorite.
 
Well, it always hurts a little bit when a person claims a domain for a function is $\{(x,y)\mid x^2>y\}$ but then sketches the regions for $\{(x,y)\mid x^2<y\}$
 
Don't worry. Worse calamities are yet to come.
In my 40-year teaching career, I never assigned such a problem :)
It's not a bad exercise, though, because when you get to 2D and 3D integration, you need to sketch.
 
my personal hell today was getting people to realize that the prediction "$v_f = \frac{m_A}{m_A+m_B}v_i$" meant that, if you plotted $v_f$ on the vertical axis and $v_i$ on the horizontal axis, then the graph should be a line through the origin with slope $m_A/(m_A+m_B)$
 
raises hand: @Semiclassic, sir, what is "slope"?
 
1:22 AM
agghhhhh
 
I wonder if I would be able to keep a straight face at that question. I would have to, but could I?
 
@Rithaniel: If you teach, you have to remember that your average student isn't as interested, motivated, or talented as you are.
 
What I have to remember myself is how bloody often I've done problems like these
for me it's as easy as breathing.
 
Yup, @Semiclassic. Easier, in fact.
 
Well, don't give me an ego, but fair enough.
 
1:25 AM
I tried to be a patient teacher ... and most of the time I try to be patient on MSE and in chat. But sometimes ...
 
1:46 AM
looking at the exam draft the prof gave me, I find myself frustrated.
in intro physics kinematics, there's three main equations you use: $x=x_0+v_0 t+\frac12 at^2$, $v=v_0+at$, and $v^2=v_0^2+2a (x-x_0)$
the first two I'm fine with. I sorta detest the last one, as useful as it is, because it's really just energy conservation in disguise
So having a quiz problem for which the most efficient solution is to use that formula repeatedly, rubs me the wrong way. they'll be doing plenty of energy conservation arguments in the future, so it seems counter-productive to teach them to use the emaciated special case of it
 
Have a quick question on how I can express a product in closed notation
I want to express $21 \choose 3$$18 \choose 3$$15 \choose 3$$\dots$$9 \choose 3$$6 \choose 3$$3 \choose 3$
Using Pi notation
 
well, each is of the form $\binom{3i}{3}$ for $i=1,2,3,\ldots,7$
So that makes for a pretty simple product
(though not the simplest way to express that product of binomial coefficients imo)
 
2:01 AM
Yea I was thinking more along the lines $(21 - 3k) \choose 3$ for $k$ from $0$ to $6$
 
same effect, yeah
 
Guess both work
 
just a different index of summation
 
Just never wrote anything in Pi notation but I just youtubed it and it seems just like sigma but for multiplication
 
yep
one cute thing about that product: try writing it entirely in terms of factorials and see what cancels. (don't expand the factorials)
 
2:11 AM
Would anyone mind helping my through some really basic algebra? I'm having a hard time.
me*
 
Semiclassical, yea i saw the pattern but didnt wana go through the hassle on my homework because i didnt wana risk losing marks for something I didn't "thoroughly explain"
 
point
main point is that this is what people introduce multinomial coefficients for
e.g. $\binom{21}{3,3,3,3,3,3,3}=\frac{21!}{(3!)^7}$
 
But I see that its 21! / (6^7)
Ah okay.
Never heard of multinomial coefficient before hmmm
Also, how would I write that pi notation in latex? This is what I'm getting $\displaystyle\prod_{n=0}^{6} [(21-3n) \choose 3]$
 
$\displaystyle\prod_{n=0}^{6} {(21-3n) \choose 3}$
replace the square brackets with curly ones
alternatively, use the \binom{} construction
e.g. \binom{21-3n}{3} -> $\binom{21-3n}{3}$
 
Ah okay. Thanks
 
2:24 AM
(4/9tw^−2)^−3 (4/9tw^−2)^3 I looked up the answer in the back of my book and it says that it's 1, but I don't understand. I know that this is a pathetically easy question, but I have no math background, and I'm learning on my own.

The parentheses mean that I'm supposed to multiply, but when I multiply the fractions, I get 16/81. and w^-6 subtracted by w^-6 is negative 12. I think I'm supposed to flip one of them to the denominator, but which one? Because then it would give me -6 minus -6 which would make it addition, thus 0, right? and that would leave me with just w.
Sorry if I'm not writing it clearly, I'm just confused.
 
x^-3 * x^3 = 1 regardless of what x is
 
Think about if you just looked at (4/9tw^-2) as a single chunk. It appears twice, each time with a particular power, so you can imagine everything in the parentheses as just a "thing" being raised to a particular power.
Compare to (4/9tw^−2)^−2 (4/9tw^−2)^3=(4/9tw^-2)
Or (4/9tw^−2) (4/9tw^−2)=(4/9tw^−2)^2
 
but what am I doing with the 4/9? Aren't I raising it to the -3 power?
 
Sure. You're also multiplying it by (4/9)^3 and (4/9)^-3 (4/9)^3=1
This is because (x^a)(x^b)=x^(a+b) and 3-3=0
Also, anything raised to the 0 power is 1.
 
So, what I'm doing is raising 4/9 to the -3 and to the 3, and then subtracting? Meaning, one gets 64/729-64/729?
 
2:34 AM
(4/9)^-3 is 729/64, not 64/729
 
Nope, because then you would have 0, not 1.
Basically, x^-3=(1/x)^3
 
(4/9)^(-3) = (9/4)^3 = 9^3/4^3
 
2:48 AM
I have 10 different balls I want to distribute among 3 different boxes, problem is that there are too much different combination of balls and boxes, I have no idea to approach it
 
so anyway i think there are cute tree diagrams that help you count finite topologies
and i think i got them right. but im not sure what (optimal) time complexity of enumerating a topology on n points would be
 
3:13 AM
@JoeShmo finite topological space are Alexandrov spaces, so they are determined by the specialization preorder, you can equivalently count preorders
 
yeah, this is what i do
 
 
2 hours later…
5:16 AM
@yuvrajsingh en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) look for stars and bars you can find video about it
 
5:27 AM
then do all possible box cases (using only one of the tree boxes,exactly 2 and using all the 3 boxes ) and multiply by the possible permutations for every case
 
doubt is that star and bar is applicable for identical things, it just example, actually I want to understand how to distribute distinct things among distinct people so that each get at least something
 
5:44 AM
think about it like that, if you add two identical balls to your 10 balls
and calculate all permutation then divide by two it should work
you can see the two new balls as bars who tel you in witch box your are
I mean you still have to remove many case by hand but it is doable
@yuvrajsingh I guess what you want is this en.wikipedia.org/wiki/Twelvefold_way
 
6:10 AM
@Rapa site correct me you mean there are 12 balls, so they arrange in 12¡ ways and divide by 2¡
 
yeah but don't forget you got to many solution caus you should remove the cases where box are empty and the case when the sames balls are in the same box with different order
I think for you case it is easier to just do 3^10 since you have 3 choice for every balls
then you remove the case with empty boxes
 
One more doubt like I have I have 7 chair in a row, how many was 4 student can sit no empty chair between any tow students @rapasite
The question clearly such that 4 student must sit together,
 
this one is easy cause you can translate the group of students
there is 4! way to align 4 students and for those you have (7-4+1) way to translate them
so 4*4!
 
(7-4+1) I didn't, t get
 
the a line of 4 students put them on the most left
there is 7-3 empty chairs on there right
 
6:22 AM
OK you are treating them as one
 
yes cause you said that they are together
1100 then 0110 then 0011
you translate the student along the 7 chairs
 
When I write 7c4 is the total case of sitting 4 children on 7 chairs
 
4! is the order of the students (4 choice for the first one then 3 choice for the second ,2 for 3rd and one for the last student)
 
And four children together is the subset of this case or not
@Rapa site one mi
Question
 
7c4 is call a combinaison
its mean "how many groups of 4 object can I make with a set of 7 objects"
so in this case the order does not matter
in your problem with the chairs the order matter so no it is different
got to go, try to find some combinatorics video or pictures to have an intuition of the basic formulas
 
6:44 AM
@rapasite a small question advice rolled 4 time, the number are listed, the number of different throws, such that largest number appearing in list is not 4
Total possible pairs are 6.6.6.6=1296
- where the highest number list may be 4
It mean rest of three position there are only 3 option that is 3,2,1
 
0
Q: Triple integral calculation

maths student$$ \text { Find the volume of the solid bounded by the surfaces } x=0 \text { and } y^{2}+z^{2}=4 \text { and } x+z=4 $$ The part which I am confused is what are limits of integration over here $x$ goes from $ 0 $ to $4-z$ ; $ y $ goes from $0$ to $ 4 - z^2$ and $ z $ from $0$ to $z$. Is it cor...

 
 
1 hour later…
7:57 AM
Hmm...
1. Nuking reflexivity will not do much here as you are not interested in equalities
2. You already considered what happened when you nuke connexivity, the resulting example you gave (1 < 2 > 3) is an example of a tree. Trees are important in computational sciences as they represent data structures and filing systems
3. You can also nuke transitivity. One of the ways gives you (2), the other way (anti transitivity), gives you functions such that the resulting partial order or preorder is cyclic (i.e. 1 < 2 < 3 < 1). This kind of function will fit the usual criteria to be monotone in t
anti transitivity is when a < b and b < c, a > c
obviously that cannot hold for all elements otherwise you have b > c and c < b but (b,c) can be anything
Thus, the near counterexamples I can came up with are trees with cycles conjoined in some elements such that e.g. 1 < 2 > 3 > 4 > 5 > 3 etc.
 
 
3 hours later…
11:16 AM
For $B \in M_{2}(\Bbb{Z_p})$ how does one show that $B^{q+2} = B^2$ where $q$ is the order of $GL_2(\Bbb{Z_p})$
 
11:34 AM
What is the smooth manifold structure on $\Bbb{R}$? I want to show that $x \mapsto x^3$ is a smooth function but it's inverse isn't. What are the chart/atlases on $\Bbb{R}$? The identity function an all open sets?
 
@user193319 you just need one chart, given by the identity function
consider three cases:
$B$ is invertible
$B$ is non-invertible, but diagonizable
$B$ is non-invertible, non-diagonizable
in the first case, $B^q=1$ by Lagrange's theorem
 
@MatheinBoulomenos That's the catch, I cannot use diagonizable. This was given to me in an exercise where only basic ring theory has been assumed including Lagrange's theorem
 
@SayanChattopadhyay I don't see how to do this without any linear algebra
if you're allowed to use some linear algebra, it's not hard though
 
@MatheinBoulomenos How would you do it using the diagonalizable cases?
 
@MatheinBoulomenos What exactly is a smooth manifold? I'm having trouble finding a concise definition. I know you start with a topological manifold (and I know what a topological manifold is). And you then you obtain a smooth manifold by putting a certain structure on it, which involves choosing certain open sets and maps into some Euclidean space.
 
11:44 AM
@MatheinBoulomenos Oh I think I see, nvm
 
well, the diagonizable case is easy, you can just check it for a diagonal matrix. In the non-diagonizable non-invertible case, you're going to have $B^2=0$
 
@user193319 You add this additional condition that transition maps between the different charts that you have, have to be smooth
 
Okay. So the open sets have to cover the topological manifold, call them $M_{\alpha}$. Then there are maps $\varphi_{\alpha}$ (homeomorphisms?) from $M_{\alpha}$ to some open set $U_{\alpha} \subseteq \Bbb{R}^n$ (assuming the $M$ is an $n$-dimensional topological manifold?).
 
@SayanChattopadhyay I just did it for rank 1 and 2 separately
 
Is $\varphi_{\alpha}$ called the chart, or $(M_{\alpha},\varphi_{\alpha})$?
 
11:49 AM
Rank 1 matrices can be written as a product [[a], [b]] * [1, s], so you can take this to the whatever power and refactor it as mostly products [1, s]*[[a], [b]], which are just numbers
And then it reduces to the original fact that x^p = x in your field; can actually do this over any finite field I think, not just Z/p
 
@MikeMiller that's a neat solution
 
Interesting @MikeMiller, I guess I can say that this is the least amount of linear algebra I have to use to do this.
 
 
1 hour later…
1:09 PM
@yuvrajsingh @yuvrajsingh can you clarify your dice roll, this is not that clear thx you
 
Consider a determinant with 1,2,3,...,n along its diagonal and 1s everywhere else. What is the value of the determinant?
 
(n-1)!
 
1:28 PM
@rapasite a dice is rolled 4 times
 
@schn the most obvious approach to this is just Gaussian elimination. It’s tedious but straightforward
 
@rapasite please look at my approach
 
A less labor-intensive approach is to recognize that that matrix is of the form $D+u u^\top$ where $D$ is a diagonal matrix and $u$ is a column vector of ones
At which point one can do stuff like appeal to the matrix determinant lemma
 
1:54 PM
anything special about a two by two matrix with a number and it's reciprocal along the diagonal and 0s elsewhere?
 
determinant is one, i guess
 
It's inverse is just the 2x2 matrix with the inverse in each entry.
 
and the eigenvalues are real and sitting along the diagonal?
 
yeah. which means that its inverse is what you get if you rotate it by 180 degrees.
But 2-by-2 diagonal matrices are simple enough that there's not much notable to say
@Ultradark yeah, they'll just be the number and the reciprocal
 
$ \begin{pmatrix} 1 & 0 \\1 & 1\end{pmatrix}\begin{pmatrix} 1 &1
\\ 0 & 1\end{pmatrix}=\begin{pmatrix}1 &1 \\ 1 & 2 \end{pmatrix}$

$ \begin{pmatrix} 1 & 0 & 0\\1 & 1 &0\\ 1 & 0 & 2\end{pmatrix}\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 &1 & 1\\ 1 & 2 &1\\ 1 & 1 & 3\end{pmatrix}$

$ \begin{pmatrix} 1 & 0 & 0 & 0\\1 & 1 & 0 & 0\\ 1 & 0 & 2 & 0 \\1 & 0 & 0 & 3\end{pmatrix}\begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 &1 & 1 & 1\\ 1 & 2 &1&1\\ 1 & 1 & 3&1 \\ 1 & 1 & 1 & 4\end{pmatrix}$
 
2:01 PM
nice.
 
and it's not hard to see the decomposition holds in general
but I was too lazy to come up with a good dots notation
 
for my tastes, I'd write that as $LDL^\top$ where $L$ is lower-triangular with ones everywhere and $D=\text{diag}(1,1,2,\ldots,n-1)$
simply because I like stuff to be symmetric
oh, wait. wrong characterization of $L$
I guess $L$ should have ones on the diagonal and on the first column, and zero everywhere else
 
I don't think you meant to put a $!$ there
 
that too
 
Now suppose there's a parameter in the matrix, but for all $s$ the determinant is still one. $A=\begin{pmatrix} e^s&0\\ 0&e^{-s} \end{pmatrix}$
 
2:06 PM
that's got a cute description: $A$ is the matrix exponential of $\begin{pmatrix} s & 0 \\ 0 & -s\end{pmatrix}$
(that's less interesting than it may sound: the matrix exponential of a diagonal matrix is just the matrix you get by replacing each matrix element with its exponentiation)
 
oh cool
 
where things get more interesting is when you deal with non-diagonal matrices
for instance: the matrix exponential of $\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix}$ is $$\exp\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix} = \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s\end{pmatrix}$$
 
so $\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix}$ can be thought of as a continuously varying transformation, acting on points in $\Bbb R^2?$
 
it's a family of linear transformations, parametrized by $s$
one thing to note, I guess: the mapping $x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} 0 & s \\ s & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} sx_2 \\ sx_1 \end{pmatrix} $ is pretty boring
 
$A$ is probably more interesting
 
2:14 PM
yep
 
this is more algebraic than analytic/differential-geometric, but the group $\left\{ \begin{pmatrix} t & 0 \\ 0 & t^{-1} \end{pmatrix}, t \in \Bbb R^\times \right\}$ is a maximal abelian Zariski-connected subgroup of $\mathrm{SL}_2(\Bbb R)$, i.e. an algebraic torus
 
$x\mapsto \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s \end{pmatrix} \begin{pmatrix} x_1\\ x_2\end{pmatrix} = \begin{pmatrix} x_1 \cosh s+x_2\sinh s\\ x_1 \sinh s+x_2\cosh s\end{pmatrix}$
 
the consideration of algebraic tori is pretty important for the structure theory of linear algebraic groups
 
though the real fun one is doing $M=\begin{pmatrix} 0 & s \\ -s & 0\end{pmatrix}$ and $A=\exp M = \begin{pmatrix} \cos s & \sin s \\ -\sin s & \cos s\end{pmatrix}$
since $A$ in that case is a rotation matrix
 
oh wow
 
2:18 PM
this kind of matrix exponential stuff shows up in a lot of places
it comes up when you solve linear systems of ODEs, for instance, and it's a Big Deal in QM
 
wait you're saying that $A$ is a rotation matrix?
 
yep
with $s$ being the angle you rotate through
 
I'm going to use $A$ so i can save ink
 
in the case I wrote, you'd be doing a CW rotation
(doing $-s$ would be CCW)
the matrix I wrote earlier with cosh/sinh doesn't have an interpretation as a rotation
but it does show up in special relativity, where it plays the role of a boost (i.e. changing your frame of reference to one that's moving faster/slower)
(for motion along one spatial direction in SR, anyways. in general you have 3 dimensions of space and 1 of time, so the matrix should be 4-by-4. but that's tedious to write out so w/e)
 
dumb question, what's the symmetry group for SR? I guess the subgroup of GL_4(R) that preserves the Minkowski inner product
I know $\varepsilon$ physics where $\varepsilon$ is very small
 
2:30 PM
Should be the Lorentz group?
 
ah, that's the name
 
or maybe it's the Poincare group? I forget which one is the relevant one: en.wikipedia.org/wiki/Poincar%C3%A9_group
 
I think we have an analogy O(3):Galileo group :: Lorentz group:Poincaré group if I got it correctly
 
yeah
Lorentz group is linear transformations, Poincare is affine
 
I should learn some SR, it seems cool
 
2:33 PM
there's also the whole Lorentz group vs. restricted Lorentz group business
which I think is basically the equivalent of O(3) vs. SO(3)
you might dig this bit: "The restricted Lorentz group SO+(1, 3) is isomorphic to the projective special linear group PSL(2,C) which is, in turn, isomorphic to the Möbius group, the symmetry group of conformal geometry on the Riemann sphere."
 
had a bit of SR in high school, but without 4-vectors or anything, we just derived time dilatation and length contraction and stuff like that from a few axioms
@Semiclassical oh wow that's really cool
 
which is a line of thinking which lead Penrose to twistor theory: en.wikipedia.org/wiki/Twistor_theory#Overview
which is cool mathematically even if it hasn't been physically all that productive
see the second half of this Baez page for details: math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html
on the Mobius / Lorentz relationship
"In other words, [a Lorentz transformation] transforms the celestial sphere (changes the appearance of the night sky) in exactly the same way that [the corresponding Mobius transformation] transforms the Riemann sphere."
is a pretty cool statement
 
there is some pretty cool math there which I have never touched
 
I wonder if the isomorphism between SO+(1,3) and PSL(2,C) is a shadow of some kind of "duality" result
do I have to learn electrodynamics before SR?
I'm kind of intrigued now
 
2:41 PM
wouldn't hurt, but not really. for stuff about the celestial sphere, definitely not necessary
 
thanks
I wonder if there is any physical meaning to (the image under the isomorphism) of the discrete subgroup PSL(2,Z) of PSL(2,C)
 
3:21 PM
What is the strategy for computing a determinant with 0 along the diagonal and positive entries everywhere else (in this case, a symmetric matrix)?
 
4:02 PM
10
Q: Prove that these linear programming problems are bounded by $O(k^{1/2})$

Mats GranvikProve that these linear programming problems are bounded by $O(k^{1/2})$ The expanded partial sums of the Möbius inverse of the Harmonic numbers have two out of three properties in common with this set of linear programming problems: $$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{n=1}...

 
@schn it's not obvious to me that there's a special strategy for such determinants
the strategies for generic determinants still apply, of course
 
4:30 PM
(I prefer the proof shown on 3B1B's channel because it's much more readily generalized - I bet you could solve $\sum1/n^4$ with the same techniques - but this^ is pretty cool)
(Though would you be able to do it under exam conditions? :P)
In other news!
Google has achieved quantum supremacy! Or has it?
 
4:57 PM
@Akiva DogAteMy: There are various tricks for this one (another, which I learned from Simmons's calculus book, is a double integral with change of variables — it's an exercise in my multivariable book). But I prefer Fourier analysis and/or residues, as they generalize immediately.
@schn The zeroes don't do anything special, although if the size of the matrix is arbitrary, perhaps this is a good set-up for mathematical induction.
 
I don't understand much, and others don't understand much of what I say.
 
So you're saying you lead a symmetric life, @Mats?
 
Eh I guess.
 
@Ted fourier analysis was the method I learned for this lol
 
Hello!
Hey what do you think is a good textbook for advanced calculus? with differential equations and stuff?
 
5:12 PM
@Simone Schaums Outlines Advanced Calculus is nice, at least for convergence tests for sums. I don't know/remember if it has much about differential equations.
 
For convergence tests I find Spivak to be quite satisfactory, but thank you anyway
A... ahem... firend... lent me a pdf of Rudin's and it has some cursory overview of differential equations, but only in some exercises.
 
@Simone: If you like Spivak for rigor, then most advanced calculus books will be disappointing. The ones with "differential equations and stuff" are oriented toward engineering, rather than rigorous mathematics.
Wow, how did my font turn into that?
 
Hi @Ted @ÍgjøgnumMeg
 
Can you be more specific about your goals and which "stuff" you are talking about?
hi demonic @Alessandro and @ÍgjøgnumMeg.
Oh, I pinged you for Semiclassic's Italian translation question a few days ago, @Alessandro.
 
You had already figured it out by the time I saw your ping iirc
 
5:16 PM
Well, sorta.
 
Well I would like some introductory words before delving into exercises, I also need some introduction about functions with 2 or more arguments
what about Spivak's Calculus on Manyfolds?
 
No, don't do that. It's way too terse and assumes you already know multivariable calculus, truly. You might look at my book or my YouTube lectures on multivariable calculus/analysis and linear algebra. Do NOT look at Rudin for this, either.
 
It should have multivariable calculus, idk about differential equations
ok, Thank you Ted, What's your book called?
 
You can find info in my profile. For differential equations, most texts are very engineering-style. I like Simmons's book on differential equations. He explains beautifully and has interesting applications.
 
quick question if we define integral dependence on a module over a commutative ring with identity will the integral closure still be a subring?
 
5:20 PM
@TedShifrin thanks a lot!
 
I’m looking for a conterexample
 
@William: Best to ask @Mathein or @Tobias that question.
 
@WilliamSun how do you define it?
 
An element $a\in A$ is integral over an $A$-module if $a^n+m_{n-1}a^{n-1}+...}m_0=0$
$a^n+m_{n-1}a^{n-1}+...+m_0=0$* corrected
 
But if you're working in a module, what do powers mean? Is this an algebra over the commutative ring?
 
5:24 PM
interesting
 
Nah because we define a ring to be integral over a module
 
Now I really don't understand. I mean, I studied integral ring extensions ...
 
so the power is defined in the ring...
 
@TedShifrin $a \in A$, coefficients in module
oh wait now $a^n$ doesn't make sense @WilliamSun
 
Yeah, there's no $1$ in the module.
This whole thing seems nuts to me. Paging @Mathein!
 
5:27 PM
Yeah you are right...thanks
If an $A$-module has a multiplication it must be an $A$-algebra right? Let me think about this for a sec
multiplication idetity i mean
 
5:43 PM
Hi @Ted and @Alessandro
wuz gooood
 
Ah, there's the algebra guru.
 
@MikeMiller there's one more way you can prove it without considering rank or diagonalizibility. Just by using the characteristic polynomial and splitting it into two cases of 0 and non zero determinant
And ofcourse Cayley Hamilton
 
6:00 PM
Well, perhaps it's the spectre of the algebra guru, instead.
 
It's really nice to finally see why the Dirac notation makes sense. The Riesz representation theorem.
 
Well, algebra is some spooky stuff. And now there is a spectre as well (or was that meant to be a spectrum?)
 
It's always spooked me.
But let's not speculate.
@Tobias: So is there any way what William was talking about makes sense?
 
@TedShifrin Not that immediately spring to mind, no
 
I've only heard about rings that are integral over other rings.
 
6:08 PM
yeah, same for me
 
Seems he's now posted this.
 
and now deleted
 
LOL
 
Anyway, I hardly do any algebra at the moment
 
6:14 PM
The most math I do is as part of an introductory course to machine learning that I am taking during the commute time I don't have any work to do
 
Does that sadden you? Or is it just the new chapter of your life?
 
I do miss doing math, but not that much
plenty of challenging things to do in my new job
 
Well, you still show up here, so you clearly haven't foresaken math completely :)
 
Fair :)
 
@TedShifrin True
 
6:15 PM
I mean: Why am I still here after 4 1/2 years of retirement?
 
My team lead is actually also a mathematician. He started his bachelor's degree just before I started my PhD
 
@yuvrajsingh I understand that you roll a dice 4 times and you write the result.
then what is your question?
 
@Ted because you love the denizens of this place so much
 
@ÍgjøgnumMeg You mean he loves to smack them
 
Yeah probably
lol
 
6:19 PM
(when they deserve it of course)
 
https://imgur.com/a/YI39FsP

The scholarship cohort in London, accompanied by an historian with crazy hair
mostly physicists though
and one PhD student in contemporary dance!
 
I have a question: imagine your light a Rope it burn in a time X (seconds), now if make a circle the same rope gonna burn in X/2 , is there In any "geometry or topology" a way to make you rope burn in X/3 making any shape or twist or whatever you want? If not why and how do you prove it?
 
What if you make the rope into a figure 8 and light the crossing point?
 
6:51 PM
I post another question about integral dependence...much interesting than the previous one I think. The integral closure behaves as a “weaker” closure operator. But I haven’t figure out what this observation can do though...
 
7:25 PM
@WilliamSun I mean at least you get idempotence
 
Yeah and it is kind of non-trivial. I wonder if there are some nice geometric structure inside of it..
 
7:48 PM
there are some connections to algebraic geometry I think
 
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